Đáp án:
$\begin{array}{l}
a)\dfrac{1}{{3 - \sqrt 5 }} - \dfrac{1}{{\sqrt 5 - 1}}\\
= \dfrac{{3 + \sqrt 5 }}{{{3^2} - 5}} - \dfrac{{\sqrt 5 + 1}}{{5 - 1}}\\
= \dfrac{{3 + \sqrt 5 }}{4} - \dfrac{{\sqrt 5 + 1}}{4}\\
= \dfrac{2}{4} = \dfrac{1}{2}\\
b)\dfrac{1}{{\sqrt 5 + 1}} + \dfrac{1}{{\sqrt 5 - 2}} - \dfrac{1}{{3 - \sqrt 5 }} - \sqrt 5 \\
= \dfrac{{\sqrt 5 - 1}}{{5 - 1}} + \dfrac{{\sqrt 5 + 2}}{{5 - 4}} - \dfrac{{3 + \sqrt 5 }}{{9 - 5}} - \sqrt 5 \\
= \dfrac{{\sqrt 5 - 1}}{4} + \sqrt 5 + 2 - \dfrac{{3 + \sqrt 5 }}{4} - \sqrt 5 \\
= \dfrac{{\sqrt 5 - 1 - 3 - \sqrt 5 }}{4} + 2\\
= \dfrac{{ - 4}}{4} + 2\\
= 1\\
c)\left( {\dfrac{{15}}{{\sqrt 6 + 1}} + \dfrac{4}{{\sqrt 6 - 2}} - \dfrac{{12}}{{3 - \sqrt 6 }}} \right).\left( {\sqrt 6 + 11} \right)\\
= \left( {\dfrac{{15\left( {\sqrt 6 - 1} \right)}}{{6 - 1}} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{{6 - 4}} - \dfrac{{12\left( {3 + \sqrt 6 } \right)}}{{9 - 6}}} \right).\left( {\sqrt 6 + 11} \right)\\
= \left( {3\sqrt 6 - 3 + 2\sqrt 6 + 4 - 12 - 4\sqrt 6 } \right).\left( {\sqrt 6 + 11} \right)\\
= \left( {\sqrt 6 - 11} \right)\left( {\sqrt 6 + 11} \right)\\
= 6 - {11^2}\\
= - 115
\end{array}$
